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Let $P(x)=ax^2+bx+c$ with $a≠0$ . The polynomial $P(x)$ is factorable in real numbers if we can find real numbers $m$ , $n$ , $k$ and $l$ so that $P(x)=(mx+n)(kx+l)$ .

a) Show that $P(x)$ is factorable if and only if $b^2−4ac≥0$ .

b) Show that $3x^2−5x+10$ is not factorable in real numbers.

I understand that in order for a polynomial to be factorable it's roots need to be rational numbers, and for that to happen the discriminant needs to be a perfect square. Yet I do not know how to put all of this together into a proof or how to show that this is the case. For example, is the perfect square statement common knowledge or do I need to explain that in some way aswell.

Itoz Darbien
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Dana Simon
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    Are you trying to factor over the reals or the rationals? If it's over the reals as you said, the roots do not need to be rationals. – John Krakov Sep 15 '23 at 01:51

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You're getting confused. This question is about the real numbers. Facts about the rational numbers and perfect squares are essentially irrelevant.

Anyway, you have the biconditional there: "if and only if." You simply need to show both directions of it. Here, we show only that if a quadratic is factorable over the reals, its discriminant is nonnegative.

$P(x)=(mx+n)(kx+l)=mkx^2+(ml+nk)x+nl$

The discriminant of P is then:

$(ml+nk)^2-4mknl=(ml)^2+(nk)^2-2mknl=(ml-nk)^2$

which is nonnegative over the reals.

Proceed along similar lines to show that this is not just a necessary condition, but a sufficient one.

Malady
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